Problem: The three-digit positive integer $N$ has a ones digit of 3. What is the probability that $N$ is divisible by 3? Express your answer as a common fraction.
Solution: Let $N = xy3$, where $x,y$ are digits.  Then $N$ is divisible by 3 if and only if the number $xy$ is.  But since $\frac{1}{3}$ of the two-digit integers are divisible by 3, our final probability is $\boxed{\frac{1}{3}}$.